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rainbow22

  • 2 years ago

Find the limit to 0 of.... [(e^2x)-1]/x Is it possible to do this w/out a calculator?

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  1. zzr0ck3r
    • 2 years ago
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    do you know l`hopital rule yet?

  2. rainbow22
    • 2 years ago
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    no... :(

  3. rainbow22
    • 2 years ago
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    You can't have a zero in the denominater. The answer is 2

  4. zzr0ck3r
    • 2 years ago
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    dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule

  5. LolWolf
    • 2 years ago
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    @zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's

  6. rainbow22
    • 2 years ago
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    idk, I'm just practicing for a test and this question was online...

  7. zzr0ck3r
    • 2 years ago
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    you nee to know how to tke a derivative to do this, you will lear that soon

  8. rainbow22
    • 2 years ago
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    I know how to take a derivative, but how will that help me solve it?

  9. zzr0ck3r
    • 2 years ago
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    take the derivative of the num and the the denom. then run the limit

  10. zzr0ck3r
    • 2 years ago
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    you will have (2e^2x)/1 now run the limi

  11. zzr0ck3r
    • 2 years ago
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    limit*

  12. LolWolf
    • 2 years ago
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    Oh, I've got one: Well we know: \[ e^x-1\sim x\\ x\to 0 \] So: \[ \lim_{x\to 0}\frac{e^{2x}-1}{x}=\lim_{u\to 0}\frac{e^{u}-1}{\frac{1}{2}u} \]Where \(u=2x\) Due to that: \[ \lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2 \]

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